LeetCode-327 Count of Range Sum

题目描述

Given an integer array nums, return the number of range sums that lie in [lower, upper] inclusive.
Range sum S(i, j) is defined as the sum of the elements in nums between indices i and j (i ≤ j), inclusive.

 

题目大意

求一个数组中不同范围的和满足给定条件的范围的个数，即lower <= nums[i] + nums[i + 1] + ... + nums[j] <= upper

（要求时间复杂度小于O（N²））

 

示例

E1

 

解题思路

基于归并的思想，将数组分为两部分，递归求两部分数组的符合条件的范围个数，再求两个数组之和的满足条件的范围个数。

 

复杂度分析

时间复杂度：O(Nlog（N）)

空间复杂度：O(N)

 

代码

class Solution {
public:
    int countRangeSum(vector<int>& nums, int lower, int upper) {
        int len = nums.size();
        // 需要使用long类型，防止两个数字相加超过int范围
        vector<long> sum(len + 1, 0);
        for(int i = 0; i < len; ++i)
            sum[i + 1] = (long)(sum[i] + nums[i]);
        
        return solve(sum, lower, upper, 0, len + 1);
    }
    
    int solve(vector<long>& sum, int lower, int upper, int low, int high) {
        // 如果数字只包含少于三个数字，则表示不能将数组分为两个部分以及一个中位数
        if(high - low <= 1)
            return 0;
        int mid = (low + high) / 2, m = mid, n = mid, count = 0;
        // 递归求取分为两个部分的数组的范围个数
        count = solve(sum, lower, upper, low, mid) + solve(sum, lower, upper, mid, high);
        // 求两个子数组之间的满足的范围个数
        for(int i = low; i < mid; ++i) {
            while(m < high && (long)(sum[m] - sum[i]) < (long)lower)
                ++m;
            while(n < high && (long)(sum[n] - sum[i]) <= (long)upper)
                ++n;
            count += n - m;
        }
        
        inplace_merge(sum.begin() + low, sum.begin() + mid, sum.begin() + high);
        
        return count;
    }
};